Problem

Source: Ukrainian Mathematical Olympiad 2023. Day 1, Problem 8.3

Tags: inequalities, algebra, fractional part



Positive integers $x, y$ satisfy the following conditions: $$\{\sqrt{x^2 + 2y}\}> \frac{2}{3}; \hspace{10mm} \{\sqrt{y^2 + 2x}\}> \frac{2}{3}$$ Show that $x = y$. Here $\{x\}$ denotes the fractional part of $x$. For example, $\{3.14\} = 0.14$. Proposed by Anton Trygub